2013 AMC 10A Problems/Problem 21

Revision as of 21:47, 6 August 2019 by Nafer (talk | contribs) (Solution 3)

Problem

A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?


$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850$

Solution 1

Let $x$ be the number of coins. After the $k^{\text{th}}$ pirate takes his share, $\frac{12-k}{12}$ of the original amount is left. Thus, we know that

$x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot \frac{9}{12} \cdot \frac{8}{12} \cdot \frac{7}{12} \cdot \frac{6}{12} \cdot \frac{5}{12} \cdot \frac{4}{12} \cdot \frac{3}{12} \cdot \frac{2}{12} \cdot \frac{1}{12}$ must be an integer. Simplifying, we get


$x \cdot \frac{11}{12} \cdot \frac{5}{6} \cdot \frac{1}{2}  \cdot \frac{7}{12} \cdot \frac{1}{2} \cdot \frac{5}{12} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \frac{1}{12}$. Now, the minimal $x$ is the denominator of this fraction multiplied out, obviously. We mentioned before that this product must be an integer. Specifically, it is an integer and it is the amount that the $12^{\text{th}}$ pirate receives, as he receives $\frac{12}{12} = 1 =$ all of what is remaining.

Thus, we know the denominator is canceled out, so the number of gold coins received is going to be the product of the numerators, $11 \cdot 5 \cdot 7 \cdot 5 = \boxed{\textbf{(D) }1925}$.

Solution 2 (Using the answer choices)

Solution $1$ mentioned the expression $x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot ... \cdot \frac{1}{12}$. Note that this is equivalent to $\frac{x \cdot 11!}{12^{11}}$.


We can compute the amount of factors of $2$, $3$, $5$, etc. but this is not necessary. To minimize this expression, we must take out factors of $2$ and $3$, since $12^{11}=2^{22} \cdot 3^{11}$. $11!$ has neither $22$ factors of $2$, nor $11$ factors of $3$. This means that if $11!$ contains $a$ factors of $2$, then $x$ will contain $22-a$ factors of $2$. This also holds for factors of $3$.


Thus, once simplified, the expression will have no factors of $2$. It will also have no factors of $3$.


Looking at the answer choices, there is only one answer which is not even, which is $\boxed{\textbf{(D) }1925}$.

Solution 3

We know that the 11th pirate takes $\frac{11}{12}$ of what is left from the 10th pirate, so we have the 12th pirate taking \[\frac{12}{12}\cdot(1-\frac{11}{12})=1\cdot\frac{1}{12}\] of what is left from the 10th pirate. Similarly, since the 10th pirate takes $\frac{10}{12}$ of what is left from the 9th pirate, we have the 11th pirate taking \[\frac{11}{12}\cdot(1-\frac{10}{12})=\frac{11}{12}\cdot\frac{2}{12}\] of what is left from the 9th pirate. Thus, the 12th pirate takes \[1\cdot\frac{1}{12}\cdot\frac{2}{12}\] of what is left from the 9th pirate. Repeating the method, we can find that the 12th pirate takes \[1\cdot\frac{1}{12}\cdot\frac{2}{12}\cdot\frac{3}{12}\cdot...\cdot\frac{10}{12}\cdot{11}{12}\] of what is left from the 1st pirate, or \[1\cdot\frac{1}{12}\cdot\frac{2}{12}\cdot\frac{3}{12}\cdot...\cdot\frac{10}{12}\cdot{11}{12}\cdot\frac{12}{12}\] of the total amount of coins.

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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